A Velociraptor is pursuing you...

A velociraptor is spotting you and goes after you. There is a shelter in the direction perpendicular to the line between you and the raptor when he spots you. So you run in the direction of the shelter at a constant speed v. The raptor is pursuing you (also at a constant speed w) adjusting at each moment in time t his direction to the line between you and him (pursue in the direction of the line of sight). Calculate the curve γ(t) the raptor is following. Do some calculations (taking realistic speeds into account) whether you would make it to your shelter (play with the distance to the shelter compared to your and the raptor’s running speeds). Draw a picture of the situation.

2. Relevant equations

3. The attempt at a solution

So I basically set this problem up as follows:

I made the line between myself and the creature the y-axis (vertical axis) with the distance between us initially at d. Then the line from our midpoint to the shelter was the horizontal (t) axis. I noticed my path to the shelter made a right triangle (looks like a 30-60-90 one) from the upper endpoint of the y axis to the shelter which was (t = ζ) units along the t axis. Now if the velociraptor adjusts his direction to pursue me at every instant his graph will look something like the Ln(t) graph.

So I continued with the analysis by determining the lengths of each corresponding path (my linear path and the beasts logarithmic path) of which I found...
- Mine was {(d/2)2 + ζ2}
- The killers was (after the arc length formula) √[1/(t^2)]*t*ln(t) which ran from 0 to ζ
* However because of the 0 endpoint I kept getting ∞ as an answer for the length (presumably due to the asymptotic behavior of the logarithmic function at t=0) so to try and rememdy this I set the lower bounds to be an absurdly small number (in terms of realistic timing) and set it as 10^-10. When I ended up calculating the lengths and times I ended up getting answers that were way off for the raptor (the human seemed to be fine)

So I guess my question is what way would you guys approach this or can you help me see the faults in my argument? Because the way I set it up the only real curve that I could see describing the path the raptor takes is ln(t) but when I try to use that at t=0 I run into issues. Any help is incredibly appreciated..

The idea is to play around with it ... that is the exercise.
You think the path the raptor takes is ln(t) ... it is not clear what you mean. Your stated problem is that ln(0) is undefined ... that's OK, use ln(t+a) where a is an adjustable parameter so that ln(a) is the initial point you need. However - what makes you think ln(t) shape is correct in the first place?

The raptor moves in both x and y coordinates... so you seem to have used "t" where most people would use an "x" (t usually denotes "time" or some arbitrary parameter and you track x(t) and y(t)), was there are reason for that?

You put the initial separation along the y axis - but the way you wrote it sounds like you are plotting "the separation between you and the raptor" along the y axis. Which is it?

The initial positions should not look like a specific triangle except by accident ... unless you are given numerical values for the initial positions that you did not share with us for some reason?

Basically, I think you need to clear up your notation.

Here's how I'd start out:

I'd draw out x-y axes, putting the raptor initial position at ##\vec r = (0,d)##, you at ##\vec u = (0,0)##, and your destination at ##\vec D = (L,0)## ... or whatever labels you want to use.
Your velocity is a constant ##\vec v = (v,0)## so your position is ##\vec u(t)=(vt,0)## ... see how this works?
The raptor also has a constant speed, but the velocity changes ... initially it is ##\vec w = -w(0,1)##
The distance between you and the raptor is ##d = |\vec u - \vec r|## and the direction the raptor is running in will be ##(\vec u - \vec r)/d## so it's velocity would be ## \vec w(t) = -w(\vec u(t) - \vec r(t))/d##

Notice that the shed is at ##t=L/v## on the ##t## axis? That is, if you actually made it?

Technically this is the long way around, but it does allow you to keep the concepts clear as you work with them.
What I'd also do is play about with computer simulations for the situation ... pick a short time step and plot the positions at each step.

Note: a velociraptor was about the size of a turkey, I'd probably just stand my ground and kick it... or, better, lure it to the shed so I can trap it inside I mean: what a discovery!

Wow.. what an incredibly clear and lucid explanation of the situation at hand! Also very much appreciated the humor at the end. I'd probably feel too bad if I kicked it so I might let it gnaw of me for a while. I thank you though Mr. Bridge! I can see where everything comes from and why its been set up the way it is. I just have one question for you..

I'm supposed to calculate the path (curve) the raptor takes γ(t) and also find if I make it to the shed. IIt appears we have two unknows w(t) (the velocity of the raptor at any instant) and r(t) (the position of the raptor at any instant). Now I imagine we would need to find an equation for r(t)? The reason I say this is to find the arc length of the path. And I know that w(t) is the derivative of r(t) so we could just use that in the arc length formula but w(t) involved r(t) and r(t) is still undefined other than an initial position r0 = (0,d) (but d also relies on r?) and presumably the endpoint of rf = (L,0) but I'm unsure as to how to calculate analytically the equation for r(t) that I can then use to calculate the arc length of the raptors path and compare with the speeds to see what fate holds for me.

##\gamma(t) = \big(x(t),y(t)\big)## is the parameterized curve, in Cartesian coordinates, for the path of the velociraptor ... ie, it is basically what I've been calling ##\vec r(t)##.

You are correct that there is still a lot of work to do ... you basically need the y coordinate in terms of the x coordinate, or some other coordinate system. The trouble with me telling you more though is that these sort of decisions are probably part of the point of the homework. That means you have to do it yourself.
The trick with these things is to draw on your understanding of different areas of mathematics.

Note:
...only the direction of the raptor's velocity changes, the magnitude is always the same. As the raptor gets closer, the x component gets bigger and the y component gets smaller. and you know that the velocity is always pointing from the current position of the raptor towards your current position. That's what the vector equations in my last post are saying.

Like I suggested before, have a go doing the problem numerically using a computer. Start with, say, 0.125 second intervals, and plot the resulting positions... see what happens to the plot as you make the time intervals smaller.

A velociraptor may have been the size of a turkey, but so is a pitbull... (checks: a bit below average pitbull yep) and they have sharper teeth with a stronger bite.http://dinosaurs.ff0000.com/dinosaurs/detail/velociraptor/
... I know it comes as humour but this is the sort of thinking you need for this sort of problem.
ie. you are asked about realistic speeds ... so human running speed is about 14mph ... but can "you" run that fast? What is the world record - etc.